Wronskian Differential Equation. $3y''+(6/x)y'+3e^xy = 0$ and two $y_1$, $y_2$ are two partial solutions of such that $W(y_1, y_2) \ne 0$.
(where $W(y_1, y_2) = W(x)$ is the Wronskian of $y_1$, $y_2$). If it is known that $W(1) = 2$, calculate $W(10)$.
I know that i have to find a function first but i do not know how. Also i can check if they are linearly independent but how can i calculate $W(10)$.  I can not find any solved examples. Please help.
 A: The Wronskian of $y_1$ and $y_2$ is defined as
$W(y_1, y_2) = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_2 y_1'; \tag 1$
we may easily find the derivative
$W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 2$
further progress is made using the given equation
$3y''+(6/x)y'+3e^xy = 0, \tag 3$
out of which the constant factor $3$ may be divided, leaving
$y'' + (2/x)y'+ e^xy = 0, \tag 4$
which we know $y_1$ and $y_2$ solve; thus we have
$W'$
$= y_1(-((2/x)y_2' + e^xy_2)) - y_2(-((2/x)y_1' + e^xy_1)) = -(2/x)y_1y_2' - e^xy_1y_2 + (2/x)y_2y_1' + e^xy_1y_2$
$= -(2/x)y_1y_2'+ (2/x)y_2y_1' = -(2/x)(y_1y_2' - y_2y_1') =-(2/x)W; \tag 5$
once the clutter of this equation is removed we are left with
$W' = -(2/x)W, \tag 6$
which is a case of Abel's identity; the unique solution to (6) taking the value $W(1)$ at $1$ is
$W(x) = W(1)\exp \left (-\displaystyle \int_1^x (2/s)\;ds \right); \tag 7$
we may easily evaluate the integral:
$\displaystyle \int_1^x (2/s)\;ds = 2\int_1^x (1/s)\;ds = 2(\ln x - \ln 1) = 2\ln x = \ln x^2; \tag 8$
thus,
$W(x) = W(1)\exp(-\ln x^2) = W(1)\exp(\ln x^{-2}) = W(1)x^{-2}.   \tag 9$
Now with
$W(1) = 2, \; x = 10, \tag{10}$
$W(x) = 2(10)^{-2} = 2(.01) = .02. \tag{11}$
